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\title{More Sequences (yay!)}
\date{February 18, 2022}

\begin{document}
\ \vspace{-.25in}

\maketitle

\vspace{-.15in}

\dfn{
Let $\{x_n\}$ be a bounded sequence. Define the following: \begin{align*}
a_n &= \sup\{x_k: k \geq n \} \\
b_n &= \inf\{x_k: k \geq n \} \\
\limsup_{n \to \infty} x_n &= \lim_{n \to \infty} a_n \\
\liminf_{n \to \infty} x_n &= \lim_{n \to \infty} b_n \\
\end{align*}
}

\prob{
Consider the sequence $\displaystyle \left\{ \cos\left(\frac{n\pi}{2}\right) \right\}$.
\begin{enumerate}[a)]
\item Let $\{a_n\}$ and $\{b_n \}$ be defined as above. 
Find the first few terms of both sequences.
\item Find $\displaystyle \limsup_{n \to \infty} \cos\left(\frac{n\pi}{2}\right)$ and  $\displaystyle \liminf_{n \to \infty}\cos\left(\frac{n\pi}{2}\right).$ 
\item Find a subsequence that converges to something other than $\displaystyle \limsup_{n \to \infty} \cos\left(\frac{n\pi}{2}\right)$ or  $\displaystyle \liminf_{n \to \infty}\cos\left(\frac{n\pi}{2}\right).$ 
\end{enumerate}
}

\vfill

\prob{
Consider the sequence $\displaystyle \left\{ \frac{\cos(n\pi/2)}{n} \right\}$.
\begin{enumerate}[a)]
\item Let $\{a_n\}$ and $\{b_n \}$ be defined as above. 
Find the first few terms of both sequences.
\item Find $\displaystyle \limsup_{n \to \infty} \frac{\cos(n\pi/2)}{n}$ and  $\displaystyle \liminf_{n \to \infty} \frac{\cos(n\pi/2)}{n}.$
\item Do any subsequences converge to some other value? 
\end{enumerate}
}

\vfill

\newpage

\dfn{
The sequence $\{x_n\}$ \textbf{converges} to $x \in \RR$ if $\displaystyle
\forall \varepsilon > 0\  \exists M \in \NN\ \forall n \geq M,\ \lvert x_n - x \rvert < \varepsilon$.
}

\dfn{
A sequence $\{x_n\}$ is a \textbf{Cauchy sequence} if $\displaystyle \boxed{ \forall \varepsilon > 0\ \exists M \in \NN\ \forall n,k \geq m,\ \lvert x_n - x_k \rvert < \varepsilon}$
}


\prob{
Show that $\displaystyle \left\{ \frac{n^2 - 1}{n^2} \right\}$ is a Cauchy sequence.
}


\end{document}